The volume of a Torus 
A torus $\mathscr{T}$ with the equation $$z^2 + \left( \sqrt{x^2 + y^2} - 2 \right)^2 = 1.$$
(a) Give an equation with a close line in the plane $Oxz$ where
  $\mathscr{T}$ is a surface of revolution.
(b) Write an parametric equation of $\mathscr{T}$.
(c) Find the aera of $\mathscr{T}$.
(d) Evaluate directly  $$\iint_{\mathscr{T}}\vec{F} \cdot \vec{dS},$$
  where $$\vec{F}=x\vec{i} + y\vec{j} + x\vec{k}$$ and, deduce the
  volume inside of $\mathscr{T}$.

(Oxford (2005), modified)

EDIT 01:
On the axis $Oxz$, we get $z^2+(x-2)^2=1$.
$$\vec{r(\theta)}:=\left (\cos{(\theta)}-2, \; \sin{(\theta)} \right) \quad 0 \le \theta \le 2\pi$$
 A: We now have $z^2+(r-2)^2=1$ with $x=r\cos\theta$ and $y=r\sin\theta$. Now we will parameterize $z=\sin\phi$ and $r=2+\cos\phi$ where $\phi$ now represents the angle from the $r$-axis in the $rz$-plane.
Now the complete parameterization is:
$x=(2+\cos\phi)\cos\theta$
$y=(2+\cos\phi)\sin\theta$
$z=\sin\phi$
This answers (b) for $\phi,\theta\in [0,2\pi)$.
I think for (a), you just want the circle you gave originally: $z^2+(x-2)^2=1$, and it is rotated around the $z$-axis.
For (c): we need to calculate $\iint_S dS$. So we need to calculate the magnitude of the normal vector using the above parameterization: $dS=||\vec{r}_\phi\times \vec{r}_\theta||d\phi d\theta$ and our integral becomes: 
$$\int_0^{2\pi} \int_0^{2\pi} ||\vec{r}_\phi\times \vec{r}_\theta||d\phi d\theta.$$
$$\vec{r}(\phi,\theta)=\left\langle(2+\cos\phi)\cos\theta, (2+\cos\phi)\sin\theta , \sin\phi\right\rangle$$
Thus
$$\vec{r}_\phi=
\left\langle
-\sin\phi\cos\theta, 
-\sin\phi\sin\theta , 
\cos\phi
\right\rangle$$
$$\vec{r}_\theta=
\left\langle
-(2+\cos\phi)\sin\theta, 
(2+\cos\phi)\cos\theta , 
0
\right\rangle$$
$$
\vec{r}_\phi\times \vec{r}_\theta=
\left\langle
-(2+\cos\phi)\cos\theta\cos\phi,
-(2+\cos\phi)\sin\theta\cos\phi,
-(2+\cos\phi)\sin\theta
\right\rangle
$$
There may be a minus sign error up there, but it doesn't matter because we are going to take the magnitude:
$$
||\vec{r}_\phi\times \vec{r}_\theta||=(2+\cos\phi)
$$
So the integral for the surface area of the torus is:
$$\int_0^{2\pi} \int_0^{2\pi} (2+\cos\phi)d\phi d\theta=8\pi^2.$$
For (d): you will need to check my work above, and then compute the integral:
$$\int_0^{2\pi} \int_0^{2\pi} \vec{F}(\vec{r}(\phi,\theta))\cdot(\vec{r}_\theta\times \vec{r}_\phi) d\phi d\theta$$
Note that I have reversed the order of the cross product above in order to make it an outward normal vector. This is important. Your question does not provide an orientation for the surface, so I have assumed an outward orientation which is more standard.
Finally, to deduce the volume of the torus, we compute the integral:
$$\int_0^{2\pi} \int_0^{2\pi} \int_0^1 |\vec{r}_s\cdot(\vec{r}_\theta\times \vec{r}_\phi)| ds d\phi d\theta$$
Using parameterization:
$x=(2+s\cos\phi)\cos\theta$
$y=(2+s\cos\phi)\sin\theta$
$z=s\sin\phi$.
The $s$ is included in order to let the 'radial parameter' in the $rz$-plane vary from $0$ to $1$ in order to sweep out the entire inside of the torus.
